For what value of \(k\) are the lines \(6x+ky-1=0\) and \(x-3y+2=0\) perpendicular to each other?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\)
Two lines are perpendicular when the product of their slopes is \(-1\). The line \(ax+by+c=0\) has slope \(-\frac{a}{b}\).
Slope of \(6x+ky-1=0\): \(\displaystyle -\frac{6}{k}\)
Slope of \(x-3y+2=0\): \(\displaystyle -\frac{1}{-3}=\frac{1}{3}\)
Product of the slopes: \(\displaystyle -\frac{6}{k}\times\frac{1}{3}=-\frac{2}{k}\)
\(\displaystyle -\frac{2}{k}=-1\), so \(k=2\)
Why the other options are incorrect
Option (1) \(-18\) — Incorrect. With \(k=-18\) the first slope is \(\frac{1}{3}\), equal to the second, so the lines are parallel.
Option (3) \(-2\) — Incorrect. The slopes are \(3\) and \(\frac{1}{3}\); their product is \(+1\), not \(-1\).
Option (4) \(18\) — Incorrect. The slopes are \(-\frac{1}{3}\) and \(\frac{1}{3}\); their product is \(-\frac{1}{9}\), not \(-1\).